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Magnetic Fields - Structured Questions

Q3. The diagram shows two magnets, supported on a yoke, placed on an electronic balance.

The magnets produce a uniform horizontal magnetic field in the region between them.

A copper wire DE is connected in the circuit shown above, and is clamped horizontally at right angles to the magnetic field.

Below is a diagram that shows a simplified plan view of the copper wire and magnets:

When the apparatus is assembled with the switch open, the reading on the electronic balance is set to 0.000 g. This reading changes to a positive value when the switch is closed.

(a) Tick in the box to indicate which of the following correctly describes the direction of the force acting on the wire DE due to the magnetic field when the switch is closed.

towards the left magnet  
towards the right magnet  
vertically up
vertically down  

 

The wire must act in a downward direction on the pan of the scale in order to register a reading.

By Newton's Third Law there must be an equal but opposite force acting on the wire - so it acts vertically upwards.

[1 mark]

(b) Label the poles of the magnets by putting N or S on each of the two dashed lines in the plan view diagram above.

The current flows from E to D

Wire movement is downwards (into the plane of the paper on the diagram)

We need to use Fleming's Left hand Motor Rule to determine that the field goes from right to left.

(see the diagram)

[1 mark]

(c) Define the tesla.

One tesla (1T) is defined as the field intensity generating one newton (N) of force, per ampere (A) of current per meter of conductor.

The syllabus specifies knowing definitions of several terms. Make sure you learn the by heart!

[1 mark]

(d) The magnets are 5.00 cm long.

When the current in the wire is 3.43 A the reading on the electronic balance is 0.620 g.

Assume the field is uniform and is zero beyond the length of the magnets.

Calculate the magnetic flux density between the magnets.

I = 3.43 A

m = 0.620 g - This mass reading is due to the force acting - you have to calculate the force.

F = 9.81 x 0.620 x 10-3 N

l = 5.00 x 10-2 m

B = ?

F = BIl

B = F/Il

B = 9.81 x 0.620 x 10-3/(3.43 x 5.00 x 10-2)

B = 0.036 T

 

[3 marks]

(Total 6 marks)